f and e^f have no common pole?

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February 1st 2010, 06:23 AM
ynj

f and e^f have no common pole?

Prove that an isolated sigularity of $f(z)$ cannot be a pole of $e^{f(z)}$ .
I know the key is to prove that $f$ and $e^f$ have no common pole, but how??
February 1st 2010, 08:11 PM
Bruno J.

Hello! Here is an idea.

It is easy to see that this is true if $z_0$ is an essential or removable singularity of $f(z)$ .

In the case of an $n$ -th order pole, then I claim that $z_0$ is an essential singularity of $g(z)=e^{f(z)}$ . Clearly it is not a regular point, so it suffices to show that $|g(z)|$ does not tend to infinity as $z \rightarrow z_0$ . To do this, it suffices to show that every neighbourhood of $z_0$ contains points $\zeta$ such that $\mbox{Re } f(\zeta)<M$ , for some fixed $M$ .
February 1st 2010, 08:38 PM
ynj

Quote:

Originally Posted by Bruno J.

Hello! Here is an idea.

It is easy to see that this is true if $z_0$ is an essential or removable singularity of $f(z)$ .

In the case of an $n$ -th order pole, then I claim that $z_0$ is an essential singularity of $g(z)=e^{f(z)}$ . Clearly it is not a regular point, so it suffices to show that $|g(z)|$ does not tend to infinity as $z \rightarrow z_0$ . To do this, it suffices to show that every neighbourhood of $z_0$ contains points $\zeta$ such that $\mbox{Re } f(\zeta)<M$ , for some fixed $M$ .

Hmm...Would you please give some more details about your last step?
March 4th 2010, 09:03 PM
Bruno J.

Hi Ynj!

I know this is a really late answer but I had forgotten about this problem and I remembered it today and solved it. So here it is.

Let $f(z)$ have a pole at $z_0$ . I claim that $g(z)=e^{f(z)}$ has an essential singularity at $z_0$ .

The problem is solved if we can show that there exists a sequence $\{z_n\} \rightarrow z_0$ such that $|e^{f(z_n)}|$ is bounded as $n \rightarrow \infty$ ; this is because if $z_0$ were a pole of $g(z)$ then we would have $g(z_n)\rightarrow \infty$ as $n \rightarrow \infty$ . This is equivalent to finding a sequence such that $\mbox{Re } f(z_n)$ remains bounded, since $|e^{x+iy}|=|e^x|$ . Since we know that $|f(z_n)|$ necessarily blows up to infinity as as $n \rightarrow \infty$ , it is the imaginary part which must become very big.

Consider $F(z)=\frac{1}{f(z)}$ . This is clearly holomorphic at $z_0$ and $F(z_0)=0$ . Let $N_1\supset N_2 \supset \dots$ be a sequence of bounded open neighbourhoods of $z_0$ whose intersection is $z_0$ . Then $\{F(N_n)\}$ is a sequence of neighbourhoods of $0$ ; in particular, each of the $F(N_n)$ is bounded, and thus $S_n=\{1/F(z) : z \in N_n\}$ is a sequence of neighbourhoods of infinity. So there are points of the imaginary axis in each of the $S_n$ ; pick $z_n \in N_n$ such that $1/F(z) \in i\mathbb{R}$ . Then the sequence $z_n$ converges to $z$ and $\mbox{Re } 1/F(z_n)=\mbox{Re } f(z_n) =0$ . Thus $|e^{f(z_n)}|=1$ for $n \geq 1$ and we are done.